I'm having difficulty solving the following equation for \(a\):

\(3a²x-2a³=x³ \)

Can anyone help me with the steps? Thanks very much in advance.

Guest Feb 20, 2017

#1**+5 **

Solve for a:

3 a^2 x - 2 a^3 = x^3

Subtract x^3 from both sides:

-2 a^3 + 3 a^2 x - x^3 = 0

The left hand side factors into a product with three terms:

-(a - x)^2 (2 a + x) = 0

Multiply both sides by -1:

(a - x)^2 (2 a + x) = 0

Split into two equations:

(a - x)^2 = 0 or 2 a + x = 0

Take the square root of both sides:

a - x = 0 or 2 a + x = 0

Add x to both sides:

a = x or 2 a + x = 0

Subtract x from both sides:

a = x or 2 a = -x

Divide both sides by 2:

**Answer: |a = x or a = -x/2**

Guest Feb 20, 2017

#1**+5 **

Best Answer

Solve for a:

3 a^2 x - 2 a^3 = x^3

Subtract x^3 from both sides:

-2 a^3 + 3 a^2 x - x^3 = 0

The left hand side factors into a product with three terms:

-(a - x)^2 (2 a + x) = 0

Multiply both sides by -1:

(a - x)^2 (2 a + x) = 0

Split into two equations:

(a - x)^2 = 0 or 2 a + x = 0

Take the square root of both sides:

a - x = 0 or 2 a + x = 0

Add x to both sides:

a = x or 2 a + x = 0

Subtract x from both sides:

a = x or 2 a = -x

Divide both sides by 2:

**Answer: |a = x or a = -x/2**

Guest Feb 20, 2017

#2**+5 **

\(3a^2x-2a^3 = x^3\\ -x^3 + 3a^2x - 2a^3 = 0\\ x^3 - 3a^2x + 2a^3 = 0\\ (x^3-a^3)+(3a^3-3a^2x)=0\\ (x-a)(x^2+ax+a^2)+3a^2(a-x)=0\\ (x-a)(x^2+ax+a^2)-3a^2(x-a)=0\\ (x-a)(x^2+ax-2a^2)=0\\ (x-a)(x^2-a^2+ax-a^2)=0\\ (x-a)((x-a)(x+a)+a(x-a))=0\\ (x-a)(x-a)(x+a+a)=0\\ (x-a)^2(x+2a)=0\\ x-a=0\quad \text{OR}\quad x+2a = 0\\ a = x\quad\text{OR}\quad a=-\dfrac{x}{2}\)

I have more detailed solution on the factorization :)

MaxWong
Feb 20, 2017